Loose ends of smooth manifolds

 Preliminary exam prep

Here we round up some theorems that have escaped previous roundings-up. Let $X,Y$ be smooth manifolds and $f:X\to Y$ a smooth map.

Theorem: (Inverse function theorem) If $df_p$ is invertible for some $p\in M$, then there exist $U\owns p$ and $V\owns f(p)$ connected such that $f|_U:U\to V$ is a diffeomorphism.

Corollary: (Stack of records theorem) If $\dim(X)=\dim(Y)$, then every regular value $y\in Y$ has a neighborhood $V\owns y$ such that $f^{-1}(Y)=U_1\sqcup \cdots \sqcup U_k$, where $f|_{U_i}:U_i\to V$ is a diffeomorphism.

Proof: Since $y\in Y$ is a regular value, $df_x$ is surjective for all $x\in f^{-1}(y)$. Since $\dim(X)=\dim(Y)$ and $df_x$ is linear, $df_x$ is an isomorphism, hence invertible. By the inverse function theorem, there exist $U\owns x$ and $V\owns y$ connected such that $f|_U:U\to V$ is a diffeomorphism. Before we actually apply this, we need to show that $f^{-1}(y)$ is a finite set.

First we note that by the preimage theorem, since $y$ is a regular value, $f^{-1}(y)$ is a submanifold of $X$ of dimension $\dim(X)-\dim(Y)=0$. Next, if $f^{-1}(y) = \{x_i\}$ were infinite, since $X$ is compact, there would be some limit point $p\in X$ of $\{x_i\}$. But then by continuity,
$$y = \lim_{i\to \infty}\left[f(x_i)\right] = f\left(\lim_{i\to\infty}\left[x_i\right]\right) = f(p),$$ so $p\in f^{-1}(y)$. But then either $p$ cannot be separated from other elements of $f^{-1}(y)$, meaning $f^{-1}(y)$ is not a manifold, or the sequence $\{x_i\}$ is finite in length. Hence $f^{-1}(y) = \{x_1,\dots,x_k\}$. Let $U_i\owns x_i$ and $V_i\owns y$ be the sets asserted to exist by the inverse function theorem (the $U_i$ may be assumed to be disjoint without loss of generality). Let $V = \bigcap_{i=1}^k V_i$ and $U_i' = f^{-1}(V)\cap U_i$, for which we still have $f|_{U_i'}:U_i'\to V$ a diffeomorphism. $\square$

Theorem: (Classification of manifolds) Up to diffeomorphism,

• the only 0-dimensional manifolds are collections of points,
• the only 1-dimensional manifolds are $S^1$ and $\R$, and
• the only 2-dimensional compact manifolds are $S^2\# (T^2)^{\#n}$ or $S^2\#(\R\P^2)^{\#n}$, for any $n\geqslant 0$.

Compact 2-manifolds are homeomorphic iff they are both (non)-orientable and have the same Euler characteristic. Note that
$$\chi\left(S^2\# (T^2)^{\#n}\right) = 2-2n, \hspace{1cm} \chi\left(S^2\#(\R\P^2)^{\#n}\right) = 2-n.$$
These surfaces are called orientable (on the left) and non-orientable (on the right) surfaces of genus $n$.

Theorem: (Stokes' theorem) For $X$ oriented and $\omega\in \Omega^{n-1}_X$, $\int_X d\omega = \int_{\dy X} \omega$.

Proposition: The tangent bundle $TX$ is always orientable.

Proof: Let $U,V\subset X$ with $\vp:U\to \R^n$ and $\psi:V\to \R^n$ trivializing maps, and $\psi\circ \vp^{-1}:\R^n\to \R^n$ the transition function. To show that $TX$ is always orientable, we need to show the Jacobian of the induced transition function (determinant of the derivative) on $TX$ is always non-negative. On $TU$ and $TV$, we have trivializing maps $(\varphi,d\varphi)$ and $(\psi,d\psi)$, giving a transition function
$$(\psi\circ \varphi^{-1}, d\psi \circ d\varphi^{-1}) = (\psi\circ \varphi^{-1}, d(\psi \circ \varphi^{-1})).$$
The Jacobian of this is
$$\det(d(\psi\circ \varphi^{-1}, d(\psi \circ \varphi^{-1}))) = \det(d(\psi\circ \varphi^{-1}), d(\psi \circ \varphi^{-1}))) = \det(d(\psi\circ \varphi^{-1}))\cdot \det(d(\psi \circ \varphi^{-1})) \>0,$$
and since $d(\psi\circ \varphi^{-1})\neq 0$ (as $\psi \circ \varphi^{-1}$ is a diffeomorphism, its derivative is an isomorphism), the result is always positive. $\square$

References: Lee (Introduction to smooth manifolds, Chapter 4), Guillemin and Pollack (Differential topology, Chapter 1

Covering spaces

 Preliminary exam prep

Let $X,Y$ be topological spaces.

Definition: A space $\widetilde X$ and a map $p:\widetilde X\to X$ are called a covering space of $X$ if either of two equivalent conditions hold:

• There is a cover $\{U_\alpha\}_{\alpha\in A}$ of $X$ such that $p^{-1}(U_\alpha)\cong \bigsqcup_{\beta\in B_\alpha} U_\beta$.
• Every point $x\in X$ has a neighborhood $U\subset X$ such that $p^{-1}(U) \cong \bigsqcup_{\beta\in B}U_\beta$.

We also demand that every $U_\beta$ is carried homeomorphically onto $U_\alpha$ (or $U$) by $p$, and the $U_\alpha$ (or $U$) are called evenly covered.

Some definitions require that $p$ be surjective. A universal cover of $X$ is a covering space that is universal with respect to this property, in that it covers all other covering spaces. Moreover, a cover that is simply connected is immediately a universal cover.

Remark: Every path connected (pc), locally path connected (lpc), and semi locally simply connected (slsc) space has a universal cover.

Theorem: (Lifting criterion) Let $Y$ be pc and lpc, and $\widetilde X$ a covering space for $X$. A map $f:Y\to X$ lifts to a map $\widetilde f:Y\to \widetilde X$ iff $f_*(\pi_1(Y))\subset p_*(\pi_1(\widetilde X))$.

Further, if the initial map $f_0$ in a homotopy $f_t:Y\to X$ lifts to $\widetilde f_0:Y\to \widetilde X$, then $f_t$ lifts uniquely to $\widetilde X$. This is called the homotopy lifting property. Next, we will see that path connected covers of $X$ may be classified via a correspondence through the fundamental group.

Theorem: Let $X$ be pc, lpc, and slsc. There is a bijection (up to isomorphism) between pc covers $p:\widetilde X\to X$ and subgroups of $\pi_1(X)$, described by $p_*(\pi_1(\widetilde X))$.

Example: Let $X=T^2$, the torus, with fundamental group $\Z\oplus \Z$. Below are some covering spaces of $p:\widetilde X\to X$ with the corresponding subgroups $p_*(\pi_1(\Z\oplus\Z))$.

Definition: Given a covering space $p:\widetilde X\to X$, an isomorphism $g$ of $\widetilde X$ for which $\id_X\circ p = p\circ g$, is called a deck transformation, the collection of which form a group $G(\widetilde X)$ under composition. Further, $\widetilde X$ is called normal (or regular) if for every $x\in X$ and every $\widetilde x_1,\widetilde x_2\in p^{-1}(x)$, there exists $g\in G(\widetilde X)$ such that $g(\widetilde x_1)=\widetilde x_2$.

For path connected covering spaces over path connected and locally path connected bases, being normal is equivalent to $p_*(\pi_1(\widetilde X))\leqslant  \pi_1(X)$ being normal. In this case, $G(\widetilde X)\cong \pi_1(X)/p_*(\pi_1(\widetilde X))$. This simplifies even more for $\widetilde X$ a universal cover, as $\pi_1(\widetilde X)=0$ then.

Theorem: Let $G$ be a group, and suppose that every $x\in X$ has a neighborhood $U\owns x$ such that $g(U)\cap h(U) = \emptyset$ whenever $g\neq h\in G$. Then:

• The quotient map $q:X\to X/G$ describes a normal cover of $X/G$.
• If $X$ is pc, then $G = G(X)$.

A group action satisfying the hypothesis of the previous theorem is called a covering space action.

Proposition: For any $n$-sheeted covering space $\widetilde X\to X$ of a finite CW complex, $\chi(\widetilde X) = n\chi(X)$.

References: Hatcher (Algebraic Topology, Chapter 1)

Differential 1-forms are closed if and only if they are exact

 Preliminary exam prep

The title refers to 1-forms in Euclidean $n$-space $\R^n$, for $n\geqslant 2$. This theorem is instructive to do in the case $n=2$, but we present it in general. We will use several facts, most importantly that the integral of a function $f:X\to Y$ over a curve $\gamma:[a,b]\to X$ is given by
$$\int_\gamma f\ dx_1\wedge \cdots \wedge dx_k = \int_a^b (f\circ \gamma)\ d(x_1\circ \gamma)\wedge \cdots \wedge d(x_n\circ \gamma),$$
where $x_1,\dots,x_n$ is some local frame on $X$. We will also use the fundamental theorem of calculus and one of its consequences, namely
$$\int_a^b \frac{\dy f}{\dy t}(t)\ dt = f(b)-f(a).$$

Theorem: A 1-form on $\R^n$ is closed if and only if it is exact, for $n\geqslant 2$.

Proof: Let $\omega = a_1dx_1+\cdots a_ndx_n\in \Omega^1_{\R^n}$ be a 1-form on $\R^n$. If there exists $\eta\in \Omega^0_{\R^n}$ such that $d\eta = \omega$, then $d\omega = d^2\eta = 0$, so the reverse direction is clear. For the forward direction, since $\omega$ is closed, we have
$$0 = d\omega = \sum_{i=1}^n \frac{\dy a_1}{\dy x_i}dx_i \wedge dx_1 + \cdots + \sum_{i=1}^n \frac{\dy a_n}{\dy x_n} dx_i\wedge dx_n \ \ \ \implies\ \ \ \frac{\dy a_i}{\dy x_j} = \frac{\dy a_j}{\dy x_i}\ \forall\ i\neq j.$$
Now fix some $(\textbf{x}_1,\dots,\textbf{x}_n)\in \R^n$, and define $f\in \Omega^0_{\R^n}$ by
$$f(\textbf x_1,\dots,\textbf x_n) = \int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega,$$
for $\gamma$ the composition of the paths
$$\begin{array}{r c l} \gamma_1\ :\ [0,\textbf x_1] & \to & \R^n, \\ t & \mapsto & (t,0,\dots,0), \end{array} \hspace{5pt} \begin{array}{r c l} \gamma_2\ :\ [0,\textbf x_2] & \to & \R^n, \\ t & \mapsto & (\textbf x_1,t,0,\dots,0), \end{array} \hspace{5pt}\cdots\hspace{5pt} \begin{array}{r c l} \gamma_n\ :\ [0,\textbf x_n] & \to & \R^n, \\ t & \mapsto & (\textbf x_1,\dots,\textbf x_{n-1},t). \end{array}$$
By applying the definition of a pullback and the change of variables formula (use $s=\gamma_i(t)$ for every $i$),
$$\begin{align*} \int_{\gamma(\textbf x_1,\dots,\textbf x_n)}\omega  & = \sum_{i=1}^n \int_{\gamma_i} a_1 dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n dx_n \\ & = \sum_{i=1}^n \int_{\gamma_i} a_1(x_1,\dots,x_n)\ dx_1 + \cdots + \sum_{i=1}^n \int_{\gamma_i}a_n(x_1,\dots,x_n)\ dx_n \\ & = \sum_{i=1}^n \int_0^{\textbf x_i} a_1(\gamma_i(t))\ d(x_1\circ \gamma_i)(t) + \cdots + \sum_{i=1}^n \int_0^{\textbf x_i}a_n(\gamma_i(t))\ d(x_n\circ \gamma_i)(t) \\ & = \int_0^{\textbf x_1} a_1(\gamma_1(t))\gamma'_1(t)\ dt + \cdots + \int_0^{\textbf x_n}a_n(\gamma_n(t))\gamma_n'(t)\ dt \\ & = \int_{(0,\dots,0)}^{(\textbf x_1,0,\dots,0)} a_1(s)\ ds + \cdots + \int_{(\textbf x_1,\dots,\textbf x_{n-1},0)}^{(\textbf x_1,\dots,\textbf x_n)}a_n(s)\ ds \\ & = \int_0^{\textbf x_1} a_1(s,0,\dots,0)\ ds + \cdots + \int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds. \end{align*}$$
To take the derivative of this, we consider the partial derivatives first. In the last variable, we have
$$\frac{\dy f}{\dy \textbf x_n} = \frac\dy{\dy \textbf x_n}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds = a_n(\textbf x_1,\dots,\textbf x_n) = a_n.$$
In the second-last variable, applying one of the identities from $\omega$ being closed, we have
$$\begin{align*} \frac{\dy f}{\dy \textbf x_{n-1}}  & = \frac\dy{\dy \textbf x_{n-1}}\int_0^{\textbf x_{n-1}}a_{n-1}(\textbf x_1,\dots,\textbf x_{n-2}, s,0)\ ds  +  \frac\dy{\dy \textbf  x_{n-1}}\int_0^{\textbf x_n}a_n(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\ & = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_n}{\dy \textbf x_{n-1}}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\ & = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + \int_0^{\textbf x_n}\frac{\dy a_{n-1}}{\dy s}(\textbf x_1,\dots,\textbf x_{n-1}, s)\ ds \\ & = a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1},0) + a_{n-1}(\textbf x_1,\dots,\textbf x_n) - a_{n-1}(\textbf x_1,\dots,\textbf x_{n-1}, 0) \\ & = a_{n-1}(\textbf x_1,\dots, \textbf x_n) \\ & = a_{n-1}. \end{align*}$$
This pattern continues. For the other variables we have telescoping sums, and we compute the partial derivative in the first variable as an example:
$$\begin{align*} \frac{\dy f}{\dy \textbf x_1}  & = \frac{\dy}{\dy \textbf x_1}\int_0^{\textbf x_1}a_1(s,0,\dots,0)\ ds + \sum_{i=2}^n\frac\dy{\dy \textbf x_1}\int_0^{\textbf x_i}a_i(\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\ & = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_i}{\dy \textbf x_1} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\ & = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \int_0^{\textbf x_i} \frac {\dy a_1}{\dy s} (\textbf x_1,\dots,\textbf x_{i-1}, s,0,\dots,0)\ ds \\ & = a_1(\textbf x_1,0,\dots,0) + \sum_{i=2}^n \left(a_1(\textbf x_1,\dots,\textbf x_i, 0,\dots,0) - a_1(\textbf x_1,\dots,\textbf x_{i-1},0,\dots,0)\right) \\ & = a_1(\textbf x_1,\dots,\textbf x_n) \\ & = a_1. \end{align*}$$
Hence we get that
$$df = \frac{\dy f}{\dy x_1} dx_1 + \cdots + \frac{\dy f}{\dy x_n}dx_n = a_1dx_1 + \cdots + a_ndx_n = \omega,$$
so $\omega$ is exact. $\square$

References: Lee (Introduction to smooth manifolds, Chapter 11)

More (co)homological constructions

 Preliminary exam prep

Recall a previous post (2016-09-16, "Complexes and their homology") that focused on constructing topological spaces in different ways and recovering the homology. Here we complete that task, introducing cellular homology. Recall a cell complex (or CW complex) $X$ was a sequence of skeleta $X_k$ for $k=0,\dots,\dim(X)$ consisting of $k$-cells $e^k_i$ and their attaching maps to the $(k-1)$-skeleton.

Cellular homology

Definition: The long exact sequence in relative homology for the pair $X_k,X_{k-1}$ shares terms with the long exact sequence for the pair $X_{k+1},X_k$, as well as $X_{k-1},X_{k-2}$. By letting $d_k$ be the composition of maps in different long exact sequences, for $k>1$, that make the diagram

commute, we get a complex of equivalence classes of chains
$$\cdots \to H_{k+1}(X_{k+1},X_k) \tov{d_{k+1}} H_k(X_k,X_{k-1})\tov{d_k} H_{k-1}(X_{k-1},X_{k-2})\to \cdots \to H_1(X_1,X_0)\tov{d_1} H_0(X_0) \tov{d_0} 0,$$
whose homology $H_k^{CW}(X) = \ker(d_k)/\text{im}(d_{k-1})$ is called the cellular homology of $X$. The map $d_1$ is the connecting map in the long exact sequence of the pair $X_1,X_0$, and $d_0=0$.

This seems quite a roundabout way of defining homology groups, but it turns out to be very useful. Note that for $k=1$, the map $d_1$ is the same as for a simplicial complex, hence

Theorem: In the context above,

1. for $k\>0$, $H^{CW}_k(X)\cong H_k(X)$;
2. for $k\>1$, $H_k(X_k,X_{k-1})=\Z^\ell$, where $\ell$ is the number of $k$-cells in $X$; and
3. for $k\>2$, $d_k(e^k_i) = \displaystyle\sum_j\deg(\underbrace{\dy e^k_i}_{S^{k-1}}\tov{f_{k,i}} X_{k-1}\tov{\pi} \underbrace{X_{k-1}/X_{k-1}-e^{k-1}_j}_{S^{k-1}})e^{k-1}_j$.

Example: Real projective space $\R\P^n$ has a cell decomposition with one cell in each dimension, and 2-to-1 attaching maps $\dy(e_k) =2X_{k-1}$ for $k>1$. This gives us a construction
$$X_0 = e_0, \hspace{1cm} X_1 = e_1 \bigsqcup_{\dy(e_1)=e_0} X_0, \hspace{1cm} X_2 = e_2 \bigsqcup_{\dy(e_2)=2e_1} X_1, \hspace{1cm} X_3 = e_3 \bigsqcup_{\dy(e_3)=2e_2} X_2, \dots$$ It is immediate that $d_0=d_1=0$, and for higher degrees, we have
$$d_k(e^k) = \deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1})e^{k-1}.$$
Since this is a map between spheres, we may apply local degree calculations. The first part is the 2-to-1 cover, where every point in $\R\P^{k-1}$ is covered by two points from $S^{k-1}$, one in each hemisphere. One covers it via the identity, the other via the antipodal map. As long as we choose a point not in $\R\P^{k-2}\subset \R\P^{k-1}$, the second step doesn't affect these degree calculations. The antipodal map $S^{k-1}\to S^{k-1}$ has degree $(-1)^k$, hence for $a$ the antipodal map, the composition has degree
$$\deg(S^{k-1}\to \R\P^{k-1}\to S^{k-1}) = \deg(\id_{S^{k-1}}) + \deg(a_{S^{k-1}}) = 1+(-1)^k = \begin{cases} 2 & k\text{ even}, \\ 0 & k \text{ odd.}\end{cases}$$

Products in (co)homology

Recall that an $n$-chain on $X$ is a map $\sigma:\Delta^n\to X$, where $\Delta^n=[v_0,\dots,v_n]$ is an $n$-simplex. These form the group $C_n$ of $n$-chains. An $n$-cochain is an element of $C^n = \Hom(C_n,\Z)$, though the coefficient group does not need to be $\Z$ necessarily.

Definition: The diagonal map $X\to X\times X$ induces a map on cohomology $H^*(X\times X)\to H^*(X)$, and by Kunneth, this gives a map $H^*(X)\otimes H^*(X)\to H^*(X)$, and is called the cup product.

For $a\in H^p(X)$ and $b\in H^q(X)$, representatives of the class $a$ are in $\Hom(C_p,\Z)$ and representatives of the class $b$ are in $\Hom(C_q,\Z)$, though we will conflate the notation for the class with that of a representative. Hence for a $(p+q)$-chain $\sigma$ the cup product of $a$ and $b$ acts as
$$(a\smile b)\sigma = a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot b\left(\sigma|_{[v_p,\dots,v_{p+q}]}\right).$$
Definition: The cap product combines $p$-cochains with $q$-chains to give $(q-p)$-chains, by
$$\begin{array}{r c l} \frown\ :\ H^p(X) \times H_q(X) & \to & H_{q-p}(X), \\\ (a, \sigma) & \mapsto & a\left(\sigma|_{[v_0,\dots,v_p]}\right)\cdot \sigma|_{[v_p,\dots,v_q]}. \end{array}$$
The cap product with the orientation form of an orientable manifold $X$ gives the isomorphism of Poincare duality.

Remark: Given a map $f:X\to Y$, the cup and cap products satisfy certain identities via the induced map on cohomology groups. Let $a,b\in H^*(Y)$ and $c\in H_*(X)$ be cochain and chain classes, for which
$$f^*(a\smile b) = f^*(a)\smile f^*(b), \hspace{1cm} a\frown f_*c = f_*(f^*a\frown c).$$
The first identity asserts that $f^*$ is a ring homomorphism and the second describes the commutativity of an appropriate diagram. The cup and cap products are related by the equation
$$a(b\frown \sigma) = (a\smile b)\sigma,$$ for $a\in H^p$, $b\in H^q$ and $\sigma\in C_{p+q}$.

References: Hatcher (Algebraic topology, Chapter 2.2), Prasolov (Elements of homology theory, Chapter 2)

Images of manifolds and transversality

 Preliminary exam prep

Let $X,Y$ be manifolds embedded in $\R^n$, and $f:X\to Y$ a map, with $df_x:T_xX\to T_{f(x)}Y$ the induced map on tangent spaces.

Definition: The map $f$ is a

• homeomorphism if it is continuous and has a continuous inverse, 
• diffeomorphism if it is smooth and has a smooth inverse,
• injection if $f(a)=f(b)$ implies $a=b$,
• immersion if $df_x$ is injective for all $x\in X$,
• embedding if it is an immersion and $df_x$ is a homeomorphism onto its image,
• submersion if $df_x$ is surjective for all $x\in X$.

Transversality is a mathematical relic whose only practical use is, perhaps, in classical algebraic geometry.

Definition: The manifolds $X$ and $Y$ are transverse if $T_pX\oplus T_pY \cong \R^n$ for every $p\in X\cap Y$. The map $f$ and $Y$ are transverse if $\text{im}(f)$ and $Y$ are transverse.

Note that being transverse (or transversal) is a symmetric, but not a reflexive, nor a transitive relation. Recall that a regular value of $f$ is $y\in Y$ such that $df_x:T_xX \to T_{f(x)}Y$ is surjective for all $x\in f^{-1}(y)$. If $y$ is not in the image of $f$, then $f^{-1}(y)$ is empty, so $y$ is trivially a regular value. Every value that is not a regular value is a critical value.

Theorem: (Preimage theorem) For every regular value $y$ of $f$, the subset $f^{-1}(y)\subset X$ is a submanifold of $X$ of dimension $\dim(X)-\dim(Y)$.

Now let $M$ be a submanifold of $Y$.

Corollary: If $f$ is transverse to $M$, then $f^{-1}(M)$ is a manifold, with $\codim_Y(M)=\codim_X(f^{-1}(M))$.

Theorem: (Transversality theorem) Let $\{g_s:X\to Y\ |\ s\in S\}$ be a smooth family of maps. If $g:X\times S\to Y$ is transverse to $M$, then for almost every $s\in S$ the map $g_s$ is transverse to $M$.

If we replace $f$ with $df$, and ask that it be transverse to $M$, then $df|_s$ is also transverse to $M$.

Example: Consider the map $g_s:X\to \R^n$ given by $g_s(X)=i(X)+s=X+s$, where $i$ is the embedding of $X$ into $\R^n$. Since $g(X\times \R^n)=\R^n$ and $g$ varies smoothly in both variables, we have that $g$ is transverse to $X$. Hence by the transversality theorem, $X$ is transverse to its translates $X+s$ for almost all $s\in \R^n$.

Theorem: (Sard) For $f$ smooth and $\dy Y=\emptyset$, almost every $y\in Y$ is a regular value of $f$ and $f|_{\dy X}$. Equivalently, the set of critical values of $f$ has measure zero.

Resources: Guillemin and Pollack (Differential topology, Chapters 1, 2), Lee (Introduction to smooth manifolds, Chapter 6)

Tools of homotopy

 Preliminary exam prep

Let $X,Y$ be topological spaces and $A$ a subspace of $X$. Recall that a path in $X$ is a continuous map $\gamma:I\to X$, and it is closed (or a loop), if $\gamma(0)=\gamma(1)$. When $X$ is pointed at $x_0$, we often require $\gamma(0)=x_0$, and call such paths (and similarly loops) based.

Definitions

Definition:

• $X$ is connected if it is not the union of two disjoint nonempty open sets.
• $X$ is path connected if any two points in $X$ have a path connecting them, or equivalently, if $\pi_0(X)=0$.
• $X$ is simply connected if every loop is contractible, or equivalently, if $\pi_1(X)=0$.
• $X$ is semi-locally simply connected if every point has a neighborhood whose inclusion into $X$ is $\pi_1$-trivial.

Path connectedness and simply connectedness have local variants. That is, for $P$ either of those properties, a space is locally $P$ if for every point $x$ and every neighborhood $U\owns x$, there is a subset $V\subset U$ on which $P$ is satisfied.

Remark: In general, $X$ is $n$-connected whenever $\pi_r(X)=0$ for all $r\leqslant n$. Note that 0-connected is path connected and 1-connected is simply connected and connected. Also observe that the suspension of path connected space is simply connected.

Definition:

• A retraction (or retract) from $X$ to $A$ is a map $r:X\to A$ such that $r|_A = \id_A$.
• A deformation retraction (or deformation retract) from $X$ to $A$ is a family of maps $f_t:X\to X$ continuous in $t,X$ such that $f_0 = \id_X$, $f_1(X) = A$, and $f_t|_A = \id_A$ for all $t$.
• A homotopy from $X$ to $Y$ is a family of maps $f_t:X\to Y$ continuous in $t,X$.
• A homotopy equivalence from $X$ to $Y$ is a map $f:X\to Y$ and a map $g:Y\to X$ such that $g\circ f \simeq \id_X$ and $f\circ g \simeq \id_Y$.

Definition: A pair $(X,A)$, where $A\subset X$ is a closed subspace, is a good pair, or has the homotopy extension property (HEP), if any of the following equivalent properties hold:

• there exists a neighborhood $U\subset X$ of $A$ such that $U$ deformation retracts onto $A$,
• $X\times \{0\}\cup A\times I$ is a retract of $X\times I$, or
• the inclusion $i:A\hookrightarrow X$ is a cofibration.

In some texts such a pair $(X,A)$ is called a neighborhood deformation retract pair, and HEP is reserved for any map $A\to X$, not necessarily the inclusion, that is a cofibration. For more on cofibrations, see a previous blog post (2016-07-31, "(Co)fibrations, suspensions, and loop spaces").

Definition: There is a functor $\pi_1:\text{Top}_*\to \text{Grp}$ called the fundamental group, that takes a pointed topological space $X$ to the space of all pointed loops on $X$, modulo path homotopy.

This may be generalized to $\pi_n$, which takes $X$ to the space of all pointed embeddings of $S^n$.

Definition: Let $G,H$ be groups. The free product of $G$ and $H$ is the group
$$G*H = \{a_1\cdots a_n\ :\ n\in \Z_{\>0}, a_i\in G\text{ or }H, a_i\in G(H)\implies a_{i+1}\in H(G)\},$$
with group operation concatenation, and identity element the empty string $\emptyset$. We also assume $e_Ge_H=e_He_G=e_G=e_H=\emptyset$, for $e_G$ ($e_H$) the identity element of $G$ ($H$).

The above construction may be generalized to a collection of groups $G_1*\cdots*G_m$, where the index may be uncountable. If every $G_\alpha=\Z$ (equivalently, has one generator), then $*_{\alpha\in A} G_\alpha$ is called the free group on $|A|$ generators.

Theorems

Theorem: (Borsuk-Ulam) Every continuous map $S^n\to \R^n$ takes a pair of antipodal points to the same value.

Theorem: (Ham Sandwich theorem) Let $U_1,\dots,U_n$ be bounded open sets in $\R^n$. There exists a hyperplane in $\R^n$ that divides each of the open sets $U_i$ into two sets of equal volume.

Volume is taken to be Lebsegue measure. The Ham sandwich theorem is an application of Borsuk-Ulam (see Terry Tao's blog post for more).

Theorem: If $X$ and $Y$ are path-connected, then $\pi_1(X\times Y)\cong \pi_1(X)\times \pi_1(Y)$.

Now suppose that $X = \bigcup_\alpha A_\alpha$ is based at $x_0$ with $x_0\in A_\alpha$ for all $\alpha$. There are natural inclusions $i_\alpha:A_\alpha\to X$ as well as $j_\alpha:A_\alpha\cap A_\beta \to A_\alpha$ and $j_\beta:A_\alpha\cap A_\beta \to A_\beta$.

Both $i_\alpha$ and $j_\alpha$ induce maps on the fundamental group, each (and all) of the $i_{\alpha*}:\pi_1(A_\alpha)\to \pi_1(X)$ extending to a map $\Phi:*_\alpha \pi_1(A_\alpha)\to \pi_1(X)$.

Theorem: (van Kampen)

• If $A_\alpha\cap A_\beta$ is path-connected, then $\Phi$ is a surjection. 
• If $A_\alpha\cap A_\beta\cap A_\gamma$ is path connected, then $\ker(\Phi) = \langle j_{\alpha*}(g)(j_{\beta*}(g))^{-1}\ |\ g\in \pi_1(A_\alpha\cap A_\beta,x_0)\rangle$.

As a consequence, if triple intersections are path connected, then $\pi_1(X) \cong *_\alpha A_\alpha /\ker(\Phi)$. Moreover, if all double intersections are contractible, then $\ker(\Phi)=0$ and $\pi_1(X)\cong *_\alpha A_\alpha$.

Proposition: If $\pi_1(X)=0$ and $\widetilde H_n(X)=0$ for all $n$, then $X$ is contractible.

References: Hatcher (Algebraic topology, Chapter 1), Tao (blog post "The Kakeya conjecture and the Ham Sandwich theorem")

Explicit pushforwards and pullbacks

 Preliminary exam prep

Here we consider a map $f:M\to N$ between manifolds of dimension $m$ and $n$, respectively, and the maps that it induces. Let $p\in M$ with $x_1,\dots,x_m$ a local chart for $U\owns p$ and $y_1,\dots,y_n$ a local chart for $V\owns f(p)$. Induced from $f$ are the differential (or pushforward) $df$ and the pullback $df^*$, which are duals of each other:
$$\begin{array}{r c l} df_p\ :\ T_p M & \to & T_{f(p)}N \\[10pt] df\ :\ TM & \to & TN \\ \alpha & \mapsto & (\beta\mapsto \alpha(\beta\circ f)) \\[10pt]\\\\ \end{array} \hspace{1cm} \begin{array}{r c l} df^*_p\ :\ T_{f(p)}^* N & \to & T_p^*M\\[10pt] df^*\ :\ T^*N & \to & T^*M \\ \omega & \mapsto & \omega\circ f\\[10pt] \bigwedge ^k T^*N & \to & \bigwedge^k T^*M\\ \omega\ dy_1\wedge\cdots \wedge dy_k & \mapsto & (\omega \circ f)\ d(y_1\circ f)\wedge \cdots \wedge d(y_k\circ f) \end{array}$$
These maps may be described by the diagram below.

Example: For example, consider the map $f:\R^3\to \R^3$ given by $f(x,y,z) = (x-y,3z^2,xz+yz)$, with the image having coordinates $(u,v,w)$. With elements
$$2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\in TM, \hspace{2cm} 2uv+\sqrt w-5\in C^\infty(N), \hspace{2cm} \cos(uv)\in T^*N,$$
we have
$$\begin{align*} df_p\left(2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\right)(2uv+\sqrt w-5) & = \left(2x\frac\dy{\dy x} - 5z\frac\dy{\dy y}\right)\left(6(x-y)z^2+\sqrt{xz+yz}-5\right)(p),\\ df_p^*\left( \cos(uv)\right) & = \cos((x-y)3z^2), \\ \left(\textstyle\bigwedge^2 df_p^*\right)(\cos(uv)du\wedge dw) & = \cos((x-y)3z^2)d(3z^2)\wedge d(xz+yz) \\ & = \cos((x-y)3z^2)\left(-6z^2\ dx\wedge dz -6z^2\ dy \wedge dz\right). \end{align*}$$